How to find indices of groups in JavaScript regular expressions match?

Question

When I write a regular expression like:

var m = /(s+).*?(l)[^l]*?(o+)/.exec(\"this is hello to you\");
console.log(m);

I get a match object

Answer 1

For global regex you want to match only fragments and iterate so first solution won't work. This is a 30 min solution based on indexOf and sums that work for this case:

https://codepen.io/cancerberoSgx/pen/qYwjjz?editors=0012#code-area

!function () {
  const regex = /\/\*\*\*@\s*([^@]+)\s*(@\*\*\*\/)/gim
  const exampleThatMatch = `
    /***@
    debug.print('hello editor, simpleNode kind is ' +
    arg.simpleNode.getKindName())
    @***/

    const a = 1 //user

    /***@
    debug.print(arg.simpleNode.getParent().getKindName())
    @***/
    `
  const text = exampleThatMatch 
  function exec(r, s) {
    function indexOfGroup(match, n) {
      var ix = match.index;
      for (var i = 1; i < n; i++)
        ix += match[i].length;
      return ix;
    }
    let result
    let lastMatchIndex = 0
    const matches = []
    while ((result = regex.exec(text))) {
      const match = []
      lastMatchIndex = text.indexOf(result[0], lastMatchIndex)
      let relIndex = 0 
      for (let i = 1; i < result.length; i++) {
        relIndex = text.indexOf(result[i], relIndex)
        match.push({ value: result[i], start: relIndex, end: relIndex + result[i].length })
      }
      matches.push(match)
    }
    return matches
  }
  const groupsWithIndex = exec(regex, text)
  console.log({RESULT: groupsWithIndex })
  // now test - let's remove everything else but matched groups 
  let frag = '' , sep = '\n#######\n'
  groupsWithIndex.forEach(match => match.forEach(group => {
    frag += text.substring(group.start, group.end) + sep
  }))
  console.log('The following are only the matched groups usign the result and text.substring just to verify it works OK:', '\n'+sep)
  console.log(frag)
}()

And just in case here is the typescript:

https://codepen.io/cancerberoSgx/pen/yjrXxx?editors=0012

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